6. Theorem of Ceva, Menelaus and Van Aubel.

By hypothesis, we have
Thus
β( |XB| C|
) + γ(|Y
|XA| |Y A| ) = (β + γ)(|A1M ′ |
|M ′ A| ).
β( |XB| C|
) + γ(|Y ) = α.
|XA| |Y A|
|A1M|
|AM|
= α
β + γ ,
and so M and M ′ coincide. Thus M must lie on the line segment XY.
Corollary 1 If G is the centroid of the triangle ABC and so α = β =
γ = 1, then G belongs to the line segment XY if and only if
|XB| |Y C|
+
|XA| |Y A|
= 1.
Corollary 2 If I is the incentre of the triangle ABC then the values
of α, β and γ are given in terms of the sidelengths of the triangle as
Thus I belongs to XY if and only if
α = a, β = b and γ = c.
b( |XB| C|
) + c(|Y ) = a.
|XA| |Y A|
Corollary 3 If H is the orthocentre of the triangle ABC then the
ratios on the sides are given by
α = tan( A), β = tan( B) and γ = tan( C.)
Then we get that H belongs to the line segment XY if and only if
(tan( B))( |XB|
|XA| ) + (tan( |Y C|
C))(
|Y A| ) = tan( A).
We also get the following result which was a question on the 2006 Irish
Invervarsity Mathematics Competition.
Theorem 5 Let ABC is a triangle and let X and Y be points on the
sides AB and AC respectively such that the line segment XY bisects the area
of ABC and the points X and Y bisects the perimeter (Figure 9). Then the
incentre I belongs to the line segment XY .
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